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2 years ago

A car accelerates from rest at a constant rate of 2m/s2 for 5s. what is the speed of a car at the end of that time?

1 answer:

4 0

Good afternoon.

We have:

$\mathsf{V_0 = 0}\\ \mathsf{a = 2 \ m/s^2}\\ \mathsf{t = 5 \ s}$

The function of velocity:

$\mathsf{V = V_0+at}\\ \\ \mathsf{V = 0 + 2t}\\ \\ \mathsf{V = 2t}$

For t = 5 s:

$\mathsf{V = 2\cdot 5}\\ \\ \boxed{\mathsf{V = 10 \ m/s}}$

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3 years ago

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Which of the following effects is produced by the high surface tension of water? Which of the following effects is produced by t

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Answer:

A water strider can walk along the surface of earth due to the surface tension of water. 

Explanation:

Fluids have a tendency to shrink to minimum possible surface area and this is called surface tension. It usually occurs due to the greater force of cohesion between molecules of same substance when compared to adhesive force between molecules of different substances. Objects with greater densities can float along water surface due to the role played by surface tension. 

When insects walk along the water surface they are pulled down due to gravity. But the force of attraction between the legs of the insect and water molecules is minimal. Thus the surface tension would always tend to maintain the flatness of water overcoming the push by the legs of the strider.

When the insect’s weight pulls it down , the surface tension pushes it upwards overcoming this force of gravity. This is how water striders move along the surface of water.

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2 years ago

A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85

artcher [175]

Answer:

$v = 10.75\,\frac{m}{s}$

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

$(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )$

The velocity of the bowling ball after the collision is:

$v = 10.75\,\frac{m}{s}$

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2 years ago

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A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re

marusya05 [52]

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

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3 years ago

____________occurs when an object travels in a curved path.

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Answer:

Acceleration

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2 years ago